cho 3 so a,b,c thoa man 0<a<b<c<1. tim gia tri lon nhat cua bieu thuc B=(a+b+c+3)[1/(a+1)+1/(b+1)+1/(c+1)]
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cho ba chu so a,b,c thoa man 0<a<b<c
cho cac so thuc a,b,c thoa man a+b+c=0 Chung minh ab+bc+ca<0
Cho 3 so a,b,c khac 0 va doi mot khac nhau thoa man a^2.(b+c)=b^2.(a+c)=2015 Tinh c^2.(a+b)
cho 3 so a,b,c la so nguyen . Trong do co 1 so nguyen am , 1 so nguyen duong va 1 so bang 0 , thoa man IaI=b^2.(b-c) . Hoi a,b,c thuoc loai so nao
tim 3 so a,b,c khac nhau va khac 0 thoa man : a/(b+c) = b/(a+c) = c/(a+b)
cho cac so a, b, c thoa man: a(a-b)+b(b-c)+c(c-a)=0
tim gia tri nho nhat cua bieu thuc: A=a^3+b^3+c^3-3abc+3ab-3c+5
cho 3 chu so a,b,c thoa man 0<a<b<c .goi A la tap hop cac so co 3 chu so moi so gom ca 3 chu so a,b,c.biet rang tong cua 2 so nho nhat trong tap hop A la 488.Khi do a+b+c=.....
cho ba so a,b,c khac 0 thoa man ab+bc +ac = 0 .tinh B=bc/a2 + ca/b2 + ab/c2
\(ab+bc+ca=0\)
=> \(\frac{ab+bc+ca}{abc}=0\)
=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\)
Đặt: \(\frac{1}{a}=x;\)\(\frac{1}{b}=y;\)\(\frac{1}{c}=z\)
Ta có: \(x+y+z=0\)
=> \(x^3+y^3+z^3=3xyz\) (tự c/m, ko c/m đc ib)
hay \(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=\frac{3}{abc}\)
\(B=\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}=\frac{abc}{a^3}+\frac{abc}{b^3}+\frac{abc}{c^3}=abc.\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\)
\(=abc.\frac{3}{abc}=3\)
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cho a,b,c la cac so thuc thoa man a+2b-3b=0 va bc +2ac-3ab=o.chung minh a=b=c